5 votes

Consider the following software items: Program-$X$, Control Flow Diagram of Program-$Y$ and Control Flow Diagram of Program-$Z$ as shown below

The values of McCabe's Cyclomatic complexity of program-$X$, program-$Y$, and program-$Z$ respectively are

- 4, 4, 7
- 3, 4, 7
- 4, 4, 8
- 4, 3, 8

14 votes

Best answer

**cyclomatic complexity = no of predicate nodes +1**

No of predicate nodes = decision making nodes

so for program X , no of predicate nodes = **3 (if, while , if ) **

cyclomatic complexity of program X = 3+1=4

For program Y, no of predicate nodes = **3 ( node3, node6, node8)**

cyclomatic complexity for program Y = 3+ 1=4

For program Z , no of predicate nodes = **6 ( 3 of X followed by 3 of Y)**

cyclomatic complexity of Z = 6+1 =7

6 votes

McCabe's Cyclomatic complexity is =E-N+2P where E=no of edge, N=no of node and P= no of disconnected graph.

for program x, E=11, N=9, P=1 so MCC=E-N+2P=11-9+1*2=4

for program y, E=10, N=8, p=1 so MCC=10-8+1*2=4

for program z, E=22, N=17, p=1 so MCC= 22-17+1*2=7

so answer is A(4,4,7).

0 votes

The cyclomatic complexity of a structured program[a] is defined

with reference to the control flow graph of the program, a directed

graph containing the basic blocks of the program, with an edge

between two basic blocks if control may pass from the first to the

second. The complexity M is then defined as.

M = E − N + 2P,

where

E = the number of edges of the graph.

N = the number of nodes of the graph.

P = the number of connected components.

Source: http://en.wikipedia.org/wiki/Cyclomatic_complexity

For first program X, E = 11, N = 9, P = 1, So M = 11-9+2*1 = 4

For second program Y, E = 10, N = 8, p = 1, So M = 10-8+2*1 = 4

For Third program X, E = 22, N = 17, p = 1, So M = 22-17+2*1 = 7